SAT Sum of Solutions Formula (-b/a) and Desmos Check

The sum of solutions formula: for a quadratic equation ax² + bx + c = 0, the solutions add to -b/a. You can get the sum without solving for either root. You can also type every Desmos line on this page into the College Board version of Desmos on our Desmos on the SAT page and watch the check work.

On the SAT, the formula is the easy part. The mistakes come from reading b before the equation is in the right form, and from not knowing which roots count. This page gives the three-line proof, three worked examples built on question stems from TTA's Digital SAT question bank, a Desmos check for each, and four practice problems with answers.

The sum of solutions formula and why it works

Call the two solutions r₁ and r₂. Any quadratic with those roots can be written as a(x - r₁)(x - r₂). Expand it:

a(x - r₁)(x - r₂) = ax² - a(r₁ + r₂)x + ar₁r₂

Match this with ax² + bx + c. The x-terms give b = -a(r₁ + r₂), and the constant terms give c = ar₁r₂. Divide each by a:

• Sum of solutions: r₁ + r₂ = -b/a
• Product of solutions: r₁r₂ = c/a

Two conditions come with the formula. The equation has to read "= 0" with every term on one side before you pick out a and b. And the formula counts roots with multiplicity, so a repeated root counts twice. If a question asks you to multiply the solutions instead, use our SAT product of roots guide.

These questions are nonlinear equations in one variable, one of the question types College Board lists in the SAT's Advanced Math area.

Example 1: read the coefficients (I do)

What is the sum of the solutions to the given equation x² - 14x + 11 = 0?

Question stem from TTA's Digital SAT question bank, record 651123.

The equation already reads "= 0". Here a = 1 and b = -14, so the sum is -b/a = -(-14)/1 = 14.

The roots themselves are 7 - √38 and 7 + √38, about 0.836 and 13.164. Solving for them takes longer and invites rounding. The coefficients give the exact sum in one step.

Check it on Desmos. Type these four lines:

1  f(x) = x^2 - 14x + 11
2  r = (14 + sqrt(14^2 - 4*11))/2
3  s = (14 - sqrt(14^2 - 4*11))/2
4  r + s

Line 1 draws a parabola. Zoom out with the minus button until you see both places it crosses the x-axis, one near x = 1 and one near x = 13. Select the curve, and gray points mark those crossings. Lines 2 and 3 are the quadratic formula (typing sqrt gives a square root), and Desmos shows each value on its line. Line 4 shows 14, the same answer the formula gave.

Example 2: factor first, then use the given sum (we do)

(x - k)² = (k - 4a)(x - k)

In the given equation, a and k are constants, where k > 4a. The sum of the solutions to the equation is 3k + 37. What is the value of a?

Question stem from TTA's Digital SAT question bank, record 651165.

Step 1 (shown). Do not expand. Both sides share the factor (x - k), so move everything to one side and pull it out:

(x - k)² - (k - 4a)(x - k) = 0

(x - k)[(x - k) - (k - 4a)] = 0

Step 2 (shown). Each factor gives a root: x = k and x = 2k - 4a. The condition k > 4a makes these two roots different, because 2k - 4a minus k equals k - 4a, which is positive. So the equation has two distinct solutions, and their sum is 3k - 4a.

Step 3 (you finish). Set 3k - 4a equal to the given sum, 3k + 37, and solve for a.

Check your answer: the 3k terms cancel, leaving -4a = 37, so a = -37/4. You never needed a value for k.

Check it on Desmos. Pick any k greater than 4a = -37, such as k = 1, and use a = -37/4:

1  h(x) = (x - 1)^2 - (1 - 4(-37/4))(x - 1)

Zoom out and the graph crosses the x-axis at x = 1 and x = 39. Those add to 40, and 3(1) + 37 = 40. Try k = 0 and the crossings move to 0 and 37, which add to 3(0) + 37.

Solving for a constant from a stated condition is also the skill behind systems with infinitely many solutions. See infinitely many solutions: find k.

Example 3: collect a root from every factor

7x²(4x - 11)(4x - u) = 0

In the given equation, u is a positive constant. The sum of the solutions to the equation is 9. What is the value of u?

Question stem from TTA's Digital SAT question bank, record 651396.

The equation is already a product set equal to 0, so list the root from each factor:

• 7x² = 0 gives x = 0. It is a double root, but adding 0 twice changes nothing.
• 4x - 11 = 0 gives x = 11/4.
• 4x - u = 0 gives x = u/4.

Set the sum equal to 9: 0 + 11/4 + u/4 = 9. Multiply by 4 to get 11 + u = 36, so u = 25. That value is positive, as the question requires.

The formula agrees. Expanded, the left side is 112x⁴ - 1008x³ + 1925x², and -(-1008)/112 = 9. For any polynomial equation, the roots, counted with multiplicity, add to minus the second coefficient divided by the leading one.

Check it on Desmos.

1  g(x) = 7x^2(4x - 11)(4x - 25)
2  0 + 11/4 + 25/4

The graph meets the x-axis at 0, 2.75 and 6.25, and line 2 shows 9. Zoom in near x = 0 and you see the graph touch the axis and turn back instead of crossing it. That is what a double root looks like. For more on reading roots from a polynomial graph, see polynomial equations on Desmos.

Distinct solutions or multiplicity?

The formula counts a repeated root as many times as it repeats. (x - 3)² = 0 has one solution value, x = 3, but -b/a counts it twice: x² - 6x + 9 = 0 gives 6/1 = 6.

In Example 3 the two readings give the same answer, because the repeated root is 0. When a question says "distinct solutions" and the repeated root is not 0, they differ, and you add each value once. List the roots first, then decide.

Your turn: four SAT-style practice problems (you do)

We wrote these four for this page. Solve each one, then check on Desmos. The answers are at the end of this section.

1. What is the sum of the solutions to 3x² - 12x - 7 = 0?
2. What is the sum of the solutions to 2x(x - 5) = 3x + 9?
3. What is the sum of the distinct solutions to (x - 2)(x² - 9x + 14) = 0?
4. In the equation 2x² - (k + 1)x + 6 = 0, k is a constant. The sum of the solutions is 5. What is the value of k?

Answers

1. 4. a = 3 and b = -12, so -b/a = 12/3 = 4. Answering 12 means you forgot to divide by a.
2. 13/2, or 6.5. Expand and move everything left: 2x² - 10x - 3x - 9 = 0, so 2x² - 13x - 9 = 0 and -b/a = 13/2. Reading b = -10 from the unmoved equation gives 5, which is wrong.
3. 9. x² - 9x + 14 = (x - 2)(x - 7), so the roots are 2, 2 and 7. The distinct values are 2 and 7, which add to 9. The cubic's -b/a is 11, because it counts 2 twice.
4. 9. Here b = -(k + 1), so the sum is (k + 1)/2. Set (k + 1)/2 = 5 to get k = 9. Check: 2x² - 10x + 6 = 0 has two real roots, about 0.70 and 4.30, which add to 5.

When algebra is faster

College Board's calculator policy says some Math questions are better done without a calculator, even though you may use one. Sum-of-solutions questions fit that description. Reading -b/a takes a few seconds. Parameter questions like Examples 2 and 3 need the algebra first, and then Desmos confirms the value you found.

Mistakes to avoid

• Reading b before the equation equals 0. In practice problem 2, the unmoved equation shows -10x, but after you move 3x across, b is -13.
• Forgetting to divide by a. For 3x² - 12x - 7 = 0, the sum is 4, not 12.
• Adding solutions that do not exist. For x² - 2x + 5 = 0, -b/a is 2, but b² - 4ac = 4 - 20 = -16, so there are no real solutions to add. On Desmos, the parabola never reaches the x-axis.

Practice next

Work through the free lesson notes Advanced Math: Quadratic Equations, Notes and Practice. For more SAT-style sets, open the TTA SAT question bank.

Related guides: SAT quadratics on Desmos, find a quadratic's minimum on Desmos, and, when two equations must both hold, how to solve systems of equations on Desmos.

Checked against College Board's calculator policy and Advanced Math overview, Desmos's College Board graphing calculator and Desmos's Supported Functions article on September 26, 2026. The Test Advantage is independent of College Board and Desmos.